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# Ten highschool boys gather at the gym for a game of basketball. Two te

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Math Expert
Joined: 02 Sep 2009
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Ten highschool boys gather at the gym for a game of basketball. Two te  [#permalink]

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02 Jun 2016, 04:36
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5
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Difficulty:

95% (hard)

Question Stats:

34% (01:06) correct 66% (01:02) wrong based on 144 sessions

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Ten highschool boys gather at the gym for a game of basketball. Two teams of 5 people each will be created. How many ways are there to create these 2 teams?

(A) 90
(B) 105
(C) 126
(D) 252
(E) 525

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Posts: 254
Re: Ten highschool boys gather at the gym for a game of basketball. Two te  [#permalink]

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02 Jun 2016, 04:47
Bunuel wrote:
Ten highschool boys gather at the gym for a game of basketball. Two teams of 5 people each will be created. How many ways are there to create these 2 teams?

(A) 90
(B) 105
(C) 126
(D) 252
(E) 525

Question involves simply selecting 5 players out of 10 players.

10C5 = $$\frac{10!}{5!5!}$$ = $$\frac{10.9.8.7.6}{5!}$$ = 252

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Re: Ten highschool boys gather at the gym for a game of basketball. Two te  [#permalink]

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02 Jun 2016, 06:23
Bunuel wrote:
Ten highschool boys gather at the gym for a game of basketball. Two teams of 5 people each will be created. How many ways are there to create these 2 teams?

(A) 90
(B) 105
(C) 126
(D) 252
(E) 525

Number of ways to select 5 people out of 10= 10!/5!5!= 252

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Posts: 648
Re: Ten highschool boys gather at the gym for a game of basketball. Two te  [#permalink]

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03 Jun 2016, 03:39
Bunuel wrote:
Ten highschool boys gather at the gym for a game of basketball. Two teams of 5 people each will be created. How many ways are there to create these 2 teams?

(A) 90
(B) 105
(C) 126
(D) 252
(E) 525

Number of ways to select 5 people out of 10= 10!/5!5!= 252

I think C should be the answer here.
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Re: Ten highschool boys gather at the gym for a game of basketball. Two te  [#permalink]

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05 Jun 2016, 13:19
KS15 wrote:
Bunuel wrote:
Ten highschool boys gather at the gym for a game of basketball. Two teams of 5 people each will be created. How many ways are there to create these 2 teams?

(A) 90
(B) 105
(C) 126
(D) 252
(E) 525

Number of ways to select 5 people out of 10= 10!/5!5!= 252

I think C should be the answer here.

Please explain for you position.I'm confused to choose C
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Re: Ten highschool boys gather at the gym for a game of basketball. Two te  [#permalink]

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06 Jun 2016, 02:48
1
Bunuel wrote:
Ten highschool boys gather at the gym for a game of basketball. Two teams of 5 people each will be created. How many ways are there to create these 2 teams?

(A) 90
(B) 105
(C) 126
(D) 252
(E) 525

For the first team 5 players can be selected as 10C5 = 252 ways
For second team 5 players can be selected as 5C5= 1 way
Now we have 2 teams and the total number of ways = (252*1)/2! = 126
I have divided by 2! because order in which the teams are created does not matter.
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Joined: 29 Mar 2016
Posts: 5
Re: Ten highschool boys gather at the gym for a game of basketball. Two te  [#permalink]

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06 Jun 2016, 03:59
Bunuel wrote:
Ten highschool boys gather at the gym for a game of basketball. Two teams of 5 people each will be created. How many ways are there to create these 2 teams?

(A) 90
(B) 105
(C) 126
(D) 252
(E) 525

This is a problem of combination.
- First we need to choose $$5$$ boys from $$10$$ = $$C^5_{10}$$
- Second we need to choose $$5$$ boys from remaining $$5$$ = $$C^5_5$$ (This is because we have already picked $$5$$ boys)
- Also as the order of team selection does not matter we can simply divide the whole equation by $$2!$$ instead of $$C^2_{10}$$

$$\frac{C^5_{10} * C^5_5}{2!}$$ $$=$$ $$\frac{\frac{10!}{5!*5!}*1}{2!}$$
$$=$$ $$126$$

$$Answer = C$$
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Re: Ten highschool boys gather at the gym for a game of basketball. Two te  [#permalink]

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01 Sep 2018, 08:17
I feel there is no need to divide by 2. We are only selecting 5 players out of 6??

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Re: Ten highschool boys gather at the gym for a game of basketball. Two te  [#permalink]

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01 Sep 2018, 08:17
I feel there is no need to divide by 2. We are only selecting 5 players out of 6??

Posted from my mobile device
Intern
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Posts: 20
Re: Ten highschool boys gather at the gym for a game of basketball. Two te  [#permalink]

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01 Sep 2018, 08:17
I feel there is no need to divide by 2. We are only selecting 5 players out of 6??

Posted from my mobile device
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Re: Ten highschool boys gather at the gym for a game of basketball. Two te  [#permalink]

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01 Sep 2018, 09:12
1
a,b,c,d,e,f,g,h,i,j be the 10 players

lets say a,b,c,d,e got selected once in 10C5 case so remaining team in 5C5 is f,g,h,i,j
at some point f,g,h,i,j will also be selected in 10C5 then remaining team in 5C5 will be a,b,c,d,e.

So there is a repetition here and thus we need to divide the total possible cases by 2!.
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Re: Ten highschool boys gather at the gym for a game of basketball. Two te  [#permalink]

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04 Sep 2018, 12:21
Bunuel wrote:
Ten highschool boys gather at the gym for a game of basketball. Two teams of 5 people each will be created. How many ways are there to create these 2 teams?

(A) 90
(B) 105
(C) 126
(D) 252
(E) 525

The first team (Team A) can be created in 10C5 ways:

10! / (5! x 5!) = (10 x 9 x 8 x 7 x 6)/(5 x 4 x 3 x 2) = 3 x 2 x 7 x 6 = 6 x 42 = 252

The second team (Team B) can be selected in 5C5 = 1 way.

Since the order in which the two teams are listed (i. e., AB versus BA) does not matter, we must divide by 2!, so the total number of ways is 252/2 = 126.

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Re: Ten highschool boys gather at the gym for a game of basketball. Two te &nbs [#permalink] 04 Sep 2018, 12:21
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